Area Question 40
Question 40
- If $A_{n}$ is the area bounded by the curve $y=(\tan x)^{n}$ and the lines $x=0, y=0$ and $x=\frac{\pi}{4}$.
Then, prove that for $n>2, A_{n}+A_{n+2}=\frac{1}{n+1}$ and deduce $\frac{1}{2 n+2}<A_{n}<\frac{1}{2 n-2}$.
$(1996,3 M)$
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Solution:
- We have, $A_{n}=\int_{0}^{\pi / 4}(\tan x)^{n} d x$
Since, $0<\tan x<1$, when $0<x<\pi / 4$
We have, $0<(\tan x)^{n+1}<(\tan x)^{n}$ for each $n \in N$
$\Rightarrow \int_{0}^{\pi / 4}(\tan x)^{n+1} d x<\int_{0}^{\pi / 4}(\tan x)^{n} d x$
$\Rightarrow \quad A_{n+1}<A_{n}$
Now, for $n>2$
$$ \begin{aligned} A_{n}+A_{n+2} & =\int_{0}^{\pi / 4}\left[(\tan x)^{n}+(\tan x)^{n+2}\right] d x \ & =\int_{0}^{\pi / 4}(\tan x)^{n}\left(1+\tan ^{2} x\right) d x \end{aligned} $$
$=\int_{0}^{\pi / 4}(\tan x)^{n} \sec ^{2} x d x$
$=\frac{1}{(n+1)}(\tan x)^{n+1}$
$=\frac{1}{(n+1)}(1-0)=\frac{1}{n+1}$
Since, $\quad A_{n+2}<A_{n+1}<A_{n}$,
then $\quad A_{n}+A_{n+2}<2 A_{n}$
$\Rightarrow \quad \frac{1}{n+1}<2 A_{n}$
$\Rightarrow \quad \frac{1}{2 n+2}<A_{n}$
Also, for $n>2 A_{n}+A_{n}<A_{n}+A_{n-2}=\frac{1}{n-1}$
$$ \begin{array}{ll} \Rightarrow & 2 A_{n}<\frac{1}{n-1} \ \Rightarrow & A_{n}<\frac{1}{2 n-2} \end{array} $$
From Eqs. (i) and (ii), $\frac{1}{2 n+2}<A_{n}<\frac{1}{2 n-2}$
